In the previous problem, I showed that if if f and g are measurable functions on σ-finite measure spaces (X,A,μ), (Y,B,[itex]\nu[/itex]), and h(x,y) = f(x)g(y), then h is measurable. Also, if f ε L1(μ) and g ε L1([itex]\nu[/itex]), then h ε L1(μx[itex]\nu[/itex]) and ∫hd(μx[itex]\nu[/itex]) = ∫fdμ ∫gd[itex]\nu[/itex].
I think that I'm supposed to use these facts to solve this problem, but after multiple dead ends I'm really not sure how.

Oh, and I'm sure that Fubini's Theorem (∫hd(μx[itex]\nu[/itex]) = ∫[∫hd[itex]\nu[/itex]]dμ = ∫[∫hdμ]d[itex]\nu[/itex] will be used as well.

Never mind, I think I've got it. Since I want to solve the integral ∫g(x)dx from x=0 to x=a = ∫∫(1/t)f(t)dtdx, inner integral from t=x to t=a, outer integral from x=0 to x=a,
the function is bounded by the curves t=x, t=a, x=0, and x=a. Another way of writing this is that the function is bounded by the curves t=0, t=a, x=0, and x=t.

Thus, by Fubini's theorem, I can rewrite the integral as ∫∫(1/t)f(t)dtdx, inner integral from t=x to t=a, outer integral from x=0 to x=a as ∫∫(1/t)f(t)dxdt, inner integral from x=0 to x=t, outer integral from t=0 to t=a. Then it becomes ∫(1/t)f(t)*tdt from t=0 to t=a
= ∫f(t)dt from t=0 to t=a = ∫f(x)dx from x=0 to x=a.